E is an elliptic curve over the finite field Z/pZ. Let N= number of points on E. If N is divisible by p, show that either N=p or p=2

Question

I feel like this wants to use the Hasse bound somehow since that's really the only tool we talked about with regard to counting points on a curve, but I'm not entirely sure how to get to that conclusion.

Answer 1

As stated, your problem is slightly wrong. Let's see first what the Hasse bound gives us. Let $N=pk$ for some $k\in \mathbb N_{>0}$ be the number of points of $E$. Then $|pk-p-1|\leq 2\sqrt{p}$. Now if $k\geq 2$ we have that $|pk-p-1|\geq p-1$. But the inequality $p-1\leq 2\sqrt{p}$ doesn't hold for $p\geq 7$. Hence, if $p\geq 7$, then $k=1$ and $N=p$.

Let us check the cases $p=3,5$. If $p=5$ and $k\geq 3$, then $|pk-p-1|\geq 9$, which is bigger than $2\sqrt{5}$. Hence the only possibilities are $k=1,2$. Let $k=2$, and suppose $E$ is an elliptic curve over $\mathbb F_5$ with $10$ points. Let $y^2=f(x)$ be a weierstrass equation for $E$ with $f(x)$ a monic polynomial of degree $3$. Since $E$ has $10$ points, $f(x)$ has exactly $1$ root, so it factors as $(x-a)g(x)$ where $a\in \mathbb F_5$ and $g(x)$ is monic irreducible of degree $2$. The number of points of $E$ is given by $$10=1+\sum_{x\in \mathbb F_5}\left(\left(\frac{f(x)}{5}\right)+1\right)=6+\sum_{x\neq a}\left(\frac{(x-a)g(x)}{5}\right)$$

Here $\displaystyle \left(\frac{\cdot}{5}\right)$ is the Legendre symbol. Since $g$ was arbitrary, up to a translation we can assume it is of the form $x^2+c$. Since translations do not change irreducibility, we must have $c=2,3$. Thus it is enough to show that for every $a\in \mathbb F_5$, $$\sum_{z\neq a}\left(\frac{(x-a)(x^2+c)}{5}\right)\neq 4$$ for $c=2,3$. Let $c=2$. If $x=1$, then $1-a$ must be a non-square, so $2$ or $3$. Thus $a=-1,3$. It is easy to rule out both cases, and the same for $c=3$.

For $p=3$, one can check directly that the elliptic curve $y^2 = x^3 + x^2 + 1$ over $\mathbb F_3$ has $6$ points and the curve $y^2 = x^3 + x^2 + 2$ has $3$ points. Therefore the right claim is: if $p$ divides $N$, then $N=p$ if $p>3$ while $N\in\{p,2p\}$ if $p=3$.